
5 Solving Systems of Linear Equations5.1 Solving Systems of
Linear Equations by Graphing5.2 Solving Systems of Linear Equations
by Substitution5.3 Solving Systems of Linear Equations by
Elimination5.4 Solving Special Systems of Linear Equations5.5
Solving Equations by Graphing5.6 Linear Inequalities in Two
Variables5.7 Systems of Linear Inequalities
Fishing (p. 265)
Pets (p. 250)
Drama Club (p. 228)
Fruit Salad (p. 255)
Roofing Contractor (p. 222)
Mathematical Thinking: Mathematically proficient students can
apply the mathematics they know to solve problems arising in
everyday life, society, and the workplace.
Drama Club (p 228)
RRo ffofiing CCo tntra tctor (((p. 2222 )2)2)
y q
FiFi hshiing ((p. 26265)5)
SEE the Big Idea
FFr iuitt SSalladd ((p 25255)5)

217
Maintaining Mathematical ProficiencyMaintaining Mathematical
ProficiencyGraphing Linear Functions (A.3.C)
Example 1 Graph 3 + y = 1 — 2 x.
Step 1 Rewrite the equation in slopeintercept form.
y = 1 — 2 x − 3
Step 2 Find the slope and the yintercept.
m = 1 — 2 and b = −3
Step 3 The yintercept is −3. So, plot (0, −3).
Step 4 Use the slope to find another point on the line.
slope = rise — run
= 1 — 2
Plot the point that is 2 units right and 1 unit up from (0, −3).
Draw a line through the two points.
Graph the equation.
1. y + 4 = x 2. 6x − y = −1 3. 4x + 5y = 20 4. −2y + 12 =
−3x
Solving and Graphing Linear Inequalities (7.10.B, A.5.B)
Example 2 Solve 2x − 17 ≤ 8x − 5. Graph the solution.
2x − 17 ≤ 8x − 5 Write the inequality.
+ 5 + 5 Add 5 to each side.
2x − 12 ≤ 8x Simplify.
− 2x − 2x Subtract 2x from each side.
−12 ≤ 6x Simplify.
−12 — 6 ≤ 6x —
6 Divide each side by 6.
−2 ≤ x Simplify.
The solution is x ≥ −2.
0−5 −4 −3 −2 −1 1 32
x ≥ –2
Solve the inequality. Graph the solution.
5. m + 4 > 9 6. 24 ≤ −6t 7. 2a − 5 ≤ 13
8. −5z + 1 < −14 9. 4k − 16 < k + 2 10. 7w + 12 ≥ 2w −
3
11. ABSTRACT REASONING The graphs of the linear functions g and
h have different slopes. The value of both functions at x = a is b.
When g and h are graphed in the same coordinate plane, what happens
at the point (a, b)?
x
y2
−1
−4
42−2−4
1(0, −3)2

218 Chapter 5 Solving Systems of Linear Equations
Using a Graphing Calculator
Mathematical Mathematical ThinkingThinking
Monitoring ProgressMonitoring ProgressUse a graphing calculator
to fi nd the point of intersection of the graphs of the two linear
equations.
1. y = −2x − 3 2. y = −x + 1 3. 3x − 2y = 2y = 1 — 2 x − 3 y = x
− 2 2x − y = 2
Mathematically profi cient students select tools, including real
objects, manipulatives, paper and pencil, and technology as
appropriate, and techniques, including mental math, estimation, and
number sense as appropriate, to solve problems. (A.1.C)
Core Core ConceptConceptFinding the Point of Intersection You
can use a graphing calculator to fi nd the point of intersection,
if it exists, of the graphs of two linear equations.
1. Enter the equations into a graphing calculator.
2. Graph the equations in an appropriate viewing window, so that
the point of intersection is visible.
3. Use the intersect feature of the graphing calculator to fi nd
the point of intersection.
Using a Graphing Calculator
Use a graphing calculator to fi nd the point of intersection, if
it exists, of the graphs of the two linear equations.
y = − 1 — 2 x + 2 Equation 1
y = 3x − 5 Equation 2
SOLUTION
The slopes of the lines are not the same, so you know that the
lines intersect. Enter the equations into a graphing calculator.
Then graph the equations in an appropriate viewing window.
Use the intersect feature to fi nd the point of intersection of
the lines.
The point of intersection is (2, 1).
−6
−4
4
66
y = − x + 212
y = 3x − 5
−6
−4
4
6
IntersectionX=2 Y=1

Section 5.1 Solving Systems of Linear Equations by Graphing
219
5.1 Solving Systems of Linear Equations by Graphing
Writing a System of Linear Equations
Work with a partner. Your family opens a bedandbreakfast. They
spend $600 preparing a bedroom to rent. The cost to your family for
food and utilities is $15 per night. They charge $75 per night to
rent the bedroom.
a. Write an equation that represents the costs.
Cost, C (in dollars)
= $15 per night ⋅
Number of nights, x + $600
b. Write an equation that represents the revenue (income).
Revenue, R (in dollars)
= $75 per night ⋅
Number of nights, x
c. A set of two (or more) linear equations is called a system of
linear equations. Write the system of linear equations for this
problem.
Essential QuestionEssential Question How can you solve a system
of linear equations?
Using a Table or Graph to Solve a System
Work with a partner. Use the cost and revenue equations from
Exploration 1 to determine how many nights your family needs to
rent the bedroom before recovering the cost of preparing the
bedroom. This is the breakeven point.
a. Copy and complete the table.
b. How many nights does your family need to rent the bedroom
before breaking even?
c. In the same coordinate plane, graph the cost equation and the
revenue equation from Exploration 1.
d. Find the point of intersection of the two graphs. What does
this point represent? How does this compare to the breakeven point
in part (b)? Explain.
Communicate Your AnswerCommunicate Your Answer 3. How can you
solve a system of linear equations? How can you check your
solution?
4. Solve each system by using a table or sketching a graph.
Explain why you chose each method. Use a graphing calculator to
check each solution.
a. y = −4.3x − 1.3 b. y = x c. y = −x − 1 y = 1.7x + 4.7 y = −3x
+ 8 y = 3x + 5
x (nights) 0 1 2 3 4 5 6 7 8 9 10 11
C (dollars)
R (dollars)
APPLYING MATHEMATICS
To be profi cient in math, you need to identify important
quantities in reallife problems and map their relationships using
tools such as diagrams, tables, and graphs.
A.2.IA.3.FA.3.GA.5.C
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS

220 Chapter 5 Solving Systems of Linear Equations
5.1 Lesson What You Will LearnWhat You Will Learn Check
solutions of systems of linear equations. Solve systems of linear
equations by graphing.
Use systems of linear equations to solve reallife problems.
Systems of Linear Equationssystem of linear equations, p.
220solution of a system of linear equations, p. 220
Previouslinear equationordered pair
Core VocabularyCore Vocabullarry
Checking Solutions
Tell whether the ordered pair is a solution of the system of
linear equations.
a. (2, 5); x + y = 7 Equation 12x − 3y = −11 Equation 2
b. (−2, 0); y = −2x − 4 Equation 1y = x + 4 Equation 2
SOLUTION
a. Substitute 2 for x and 5 for y in each equation.
Equation 1 Equation 2
x + y = 7 2x − 3y = −11
2 + 5 =?
7 2(2) − 3(5) =?
−11
7 = 7 ✓ −11 = −11 ✓ Because the ordered pair (2, 5) is a
solution of each equation, it is a solution of
the linear system.
b. Substitute −2 for x and 0 for y in each equation.
Equation 1 Equation 2
y = −2x − 4 y = x + 4
0 =?
−2(−2) − 4 0 =?
−2 + 4
0 = 0 ✓ 0 ≠ 2 ✗ The ordered pair (−2, 0) is a solution of the fi
rst equation, but it is not a solution
of the second equation. So, (−2, 0) is not a solution of the
linear system.
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.comTell whether the ordered pair is a
solution of the system of linear equations.
1. (1, −2); 2x + y = 0−x + 2y = 5
2. (1, 4); y = 3x + 1y = −x + 5
READINGA system of linear equations is also called a linear
system.
A system of linear equations is a set of two or more linear
equations in the same variables. An example is shown below.
x + y = 7 Equation 1
2x − 3y = −11 Equation 2
A solution of a system of linear equations in two variables is
an ordered pair that is a solution of each equation in the
system.

Section 5.1 Solving Systems of Linear Equations by Graphing
221
Solving Systems of Linear Equations by GraphingThe solution of a
system of linear equations is the point of intersection of the
graphs of the equations.
Core Core ConceptConceptSolving a System of Linear Equations by
GraphingStep 1 Graph each equation in the same coordinate
plane.
Step 2 Estimate the point of intersection.
Step 3 Check the point from Step 2 by substituting for x and y
in each equation of the original system.
Solving a System of Linear Equations by Graphing
Solve the system of linear equations by graphing.
y = −2x + 5 Equation 1
y = 4x − 1 Equation 2
SOLUTION
Step 1 Graph each equation.
Step 2 Estimate the point of intersection. The graphs appear to
intersect at (1, 3).
Step 3 Check your point from Step 2.
Equation 1 Equation 2
y = −2x + 5 y = 4x − 1
3 =?
−2(1) + 5 3 =?
4(1) − 1
3 = 3 ✓ 3 = 3 ✓ The solution is (1, 3).
Check
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.comSolve the system of linear equations by
graphing.
3. y = x − 2 4. y = 1 — 2 x + 3 5. 2x + y = 5
y = −x + 4 y = − 3 — 2 x − 5 3x − 2y = 4
REMEMBERNote that the linear equations are in slopeintercept
form. You can use the method presented in Section 3.5 to graph the
equations.
x
y
2
42−2−4
(1, 3)y = −2x + 5
1 3)
y = 4x − 1
−1
−6
−2
6
6
y = 4x − 1y = −2x + 5
IntersectionX=1 Y=3

222 Chapter 5 Solving Systems of Linear Equations
Modeling with Mathematics
A roofi ng contractor buys 30 bundles of shingles and 4 rolls of
roofi ng paper for $1040. In a second purchase (at the same
prices), the contractor buys 8 bundles of shingles for $256. Find
the price per bundle of shingles and the price per roll of roofi ng
paper.
SOLUTION
1. Understand the Problem You know the total price of each
purchase and how many of each item were purchased. You are asked to
fi nd the price of each item.
2. Make a Plan Use a verbal model to write a system of linear
equations that represents the problem. Then solve the system of
linear equations.
3. Solve the Problem
Words 30 ⋅ Price per bundle + 4 ⋅ Price per roll
= 1040
8 ⋅ Price per bundle + 0 ⋅ Price per roll = 256 Variables Let x
be the price (in dollars) per bundle and let y be the
price (in dollars) per roll.
System 30x + 4y = 1040 Equation 1
8x = 256 Equation 2
Step 1 Graph each equation. Note that only the fi rst quadrant
is shown because x and y must be positive.
Step 2 Estimate the point of intersection. The graphs appear to
intersect at (32, 20).
Step 3 Check your point from Step 2.
Equation 1 Equation 2
30x + 4y = 1040 8x = 256
30(32) + 4(20) =?
1040 8(32) =?
256
1040 = 1040 ✓ 256 = 256 ✓ The solution is (32, 20). So, the
price per bundle of shingles is $32, and the
price per roll of roofi ng paper is $20.
4. Look Back You can use estimation to check that your solution
is reasonable. A bundle of shingles costs about $30. So, 30 bundles
of shingles and 4 rolls of roofi ng paper (at $20 per roll) cost
about 30(30) + 4(20) = $980, and 8 bundles of shingles costs about
8(30) = $240. These prices are close to the given values, so the
solution seems reasonable.
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.com 6. You have a total of 18 math and
science exercises for homework. You have
six more math exercises than science exercises. How many
exercises do you have in each subject?
Solving RealLife Problems
A$sr
S
1
2
3
8 16 240 32 x
80
160
240
320
0
y
(32, 20)
y = −7.5x + 260
x = 32

Section 5.1 Solving Systems of Linear Equations by Graphing
223
Tutorial Help in English and Spanish at
BigIdeasMath.comExercises5.1
Monitoring Progress and Modeling with MathematicsMonitoring
Progress and Modeling with MathematicsIn Exercises 3–8, tell
whether the ordered pair is a solution of the system of linear
equations. (See Example 1.)
3. (2, 6); x + y = 83x − y = 0 4. (8, 2);
x − y = 62x − 10y = 4
5. (−1, 3); y = −7x − 4y = 8x + 5
6. (−4, −2); y = 2x + 6y = −3x − 14
7. (−2, 1); 6x + 5y = −72x − 4y = −8 8. (5, −6);
6x + 3y = 124x + y = 14
In Exercises 9–12, use the graph to solve the system of linear
equations. Check your solution.
9. x − y = 4 10. x + y = 5 4x + y = 1 y − 2x = −4
x
y
−2
42
x
y
2
4
41
11. 6y + 3x = 18 12. 2x − y = −2 −x + 4y = 24 2x + 4y = 8
x
y
2
4
−2−4−6
x
y
4
2−2
In Exercises 13–20, solve the system of linear equations by
graphing. (See Example 2.)
13. y = −x + 7 14. y = −x + 4 y = x + 1 y = 2x − 8
15. y = 1 — 3 x + 2 16. y = 3 — 4 x − 4
y = 2 — 3 x + 5 y = − 1 — 2 x + 11
17. 9x + 3y = −3 18. 4x − 4y = 20 2x − y = −4 y = −5
19. x − 4y = −4 20. 3y + 4x = 3 −3x − 4y = 12 x + 3y = −6
ERROR ANALYSIS In Exercises 21 and 22, describe and correct the
error in solving the system of linear equations.
21. The solution of
the linear system x − 3y = 6 and 2x − 3y = 3 is (3, −1).
✗x
y
−1
2
2
22. The solution of
the linear system y = 2x − 1 and y = x + 1 is x = 2.
✗
x
y
2
4
42
1. VOCABULARY Do the equations 5y − 2x = 18 and 6x = −4y − 10
form a system of linear equations? Explain.
2. DIFFERENT WORDS, SAME QUESTION Consider the system of linear
equations −4x + 2y = 4 and 4x − y = −6. Which is different? Find
“both” answers.
Solve the system of linear equations. Solve each equation for
y.
Find the point of intersection of the graphs of the
equations.
Find an ordered pair that is a solution of each equation in the
system.
Vocabulary and Core Concept CheckVocabulary and Core Concept
Check

224 Chapter 5 Solving Systems of Linear Equations
Maintaining Mathematical ProficiencyMaintaining Mathematical
ProficiencySolve the literal equation for y. (Section 1.4)
34. 10x + 5y = 5x + 20 35. 9x + 18 = 6y − 3x 36. 3 — 4 x + 1 — 4
y = 5
Reviewing what you learned in previous grades and lessons
USING TOOLS In Exercises 23–26, use a graphing calculator to
solve the system of linear equations.
23. 0.2x + 0.4y = 4 24. −1.6x − 3.2y = −24 −0.6x + 0.6y = −3
2.6x + 2.6y = 26
25. −7x + 6y = 0 26. 4x − y = 1.50.5x + y = 2 2x + y = 1.5
27. MODELING WITH MATHEMATICS You have 40 minutes to exercise at
the gym, and you want to burn 300 calories total using both
machines. How much time should you spend on each machine? (See
Example 3.)
Elliptical Trainer
8 calories per minute
Stationary Bike
6 calories per minute
28. MODELING WITH MATHEMATICS You sell small and large candlesat
a craft fair. You collect $144 selling a total of 28 candles. How
many of each type of candle did you sell?
29. MATHEMATICAL CONNECTIONS Write a linear equation that
represents the area and a linear equation that represents the
perimeter of the rectangle. Solve the system of linear equations by
graphing. Interpret your solution.
6 cm
(3x − 3) cm
30. THOUGHT PROVOKING Your friend’s bank account balance (in
dollars) is represented by the equation y = 25x + 250, where x is
the number of months. Graph this equation. After 6 months, you want
to have the same account balance as your friend. Write a linear
equation that represents your account balance. Interpret the slope
and yintercept of the line that represents your account
balance.
31. COMPARING METHODS Consider the equation x + 2 = 3x − 4.
a. Solve the equation using algebra.
b. Solve the system of linear equations y = x + 2 and y = 3x − 4
by graphing.
c. How is the linear system and the solution in part (b) related
to the original equation and the solution in part (a)?
32. HOW DO YOU SEE IT? A teacher is purchasing binders for
students. The graph shows the total costs of ordering x binders
from three different companies.
1500
50
75
100
125
150
20 25 30 35
Co
st (
do
llars
)
40 45 50 x
y
Number of binders
Buying Binders
Company A
Company B
Company C
a. For what numbers of binders are the costs the same at two
different companies? Explain.
b. How do your answers in part (a) relate to systems of linear
equations?
33. MAKING AN ARGUMENT You and a friend are going hiking but
start at different locations. You start at the trailhead and walk 5
miles per hour. Your friend starts 3 miles from the trailhead and
walks 3 miles per hour.
you
your friend
a. Write and graph a system of linear equations that represents
this situation.
b. Your friend says that after an hour of hiking you will both
be at the same location on the trail. Is your friend correct? Use
the graph from part (a) to explain your answer.
$6each $4
each

Section 5.2 Solving Systems of Linear Equations by Substitution
225
5.2 Solving Systems of Linear Equations by Substitution
Using Substitution to Solve Systems
Work with a partner. Solve each system of linear equations using
two methods.
Method 1 Solve for x fi rst.Solve for x in one of the equations.
Substitute the expression for x into the other equation to fi nd y.
Then substitute the value of y into one of the original equations
to fi nd x.
Method 2 Solve for y fi rst.Solve for y in one of the equations.
Substitute the expression for y into the other equation to fi nd x.
Then substitute the value of x into one of the original
equations
to fi nd y.
Is the solution the same using both methods? Explain which
method you would prefer
to use for each system.
a. x + y = −7 b. x − 6y = −11 c. 4x + y = −1
−5x + y = 5 3x + 2y = 7 3x − 5y = −18
Essential QuestionEssential Question How can you use
substitution to solve a system of linear equations?
Writing and Solving a System of Equations
Work with a partner.
a. Write a random ordered pair with integer coordinates. One way
to do this is to use a graphing calculator. The ordered pair
generated at the right is (−2, −3).
b. Write a system of linear equations that has your ordered pair
as its solution.
c. Exchange systems with your partner and use one of the methods
from Exploration 1 to solve the system. Explain your choice of
method.
Communicate Your AnswerCommunicate Your Answer 3. How can you
use substitution to solve a system of linear equations?
4. Use one of the methods from Exploration 1 to solve each
system of linear equations. Explain your choice of method. Check
your solutions.
a. x + 2y = −7 b. x − 2y = −6 c. −3x + 2y = −10
2x − y = −9 2x + y = −2 −2x + y = −6
d. 3x + 2y = 13 e. 3x − 2y = 9 f. 3x − y = −6
x − 3y = −3 −x − 3y = 8 4x + 5y = 11
USING PRECISE MATHEMATICAL LANGUAGE
To be profi cient in math, you need to communicate precisely
with others.
randInt(5‚5‚2){2 3}
Choose tworandom integersbetween −5 and 5.
A.2.IA.5.C
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS

226 Chapter 5 Solving Systems of Linear Equations
5.2 Lesson What You Will LearnWhat You Will Learn Solve systems
of linear equations by substitution.
Use systems of linear equations to solve reallife problems.
Solving Linear Systems by SubstitutionPrevioussystem of linear
equationssolution of a system of linear equations
Core VocabularyCore Vocabullarry
Core Core ConceptConceptSolving a System of Linear Equations by
SubstitutionStep 1 Solve one of the equations for one of the
variables.
Step 2 Substitute the expression from Step 1 into the other
equation and solve for the other variable.
Step 3 Substitute the value from Step 2 into one of the original
equations and solve.
Solving a System of Linear Equations by Substitution
Solve the system of linear equations by substitution.
y = −2x − 9 Equation 1
6x − 5y = −19 Equation 2
SOLUTION
Step 1 Equation 1 is already solved for y.
Step 2 Substitute −2x − 9 for y in Equation 2 and solve for
x.
6x − 5y = −19 Equation 2
6x − 5(−2x − 9) = −19 Substitute −2x − 9 for y.
6x + 10x + 45 = −19 Distributive Property
16x + 45 = −19 Combine like terms.
16x = −64 Subtract 45 from each side.
x = −4 Divide each side by 16.
Step 3 Substitute −4 for x in Equation 1 and solve for y.
y = −2x − 9 Equation 1
= −2(−4) − 9 Substitute −4 for x.
= 8 − 9 Multiply.
= −1 Subtract.
The solution is (−4, −1).
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.comSolve the system of linear equations by
substitution. Check your solution.
1. y = 3x + 14 2. 3x + 2y = 0 3. x = 6y − 7y = −4x y = 1 — 2 x −
1 4x + y = −3
Another way to solve a system of linear equations is to use
substitution.
Check
Equation 1
y = −2x − 9
−1 =?
−2(−4) − 9
−1 = −1 ✓
Equation 2
6x − 5y = −19
6(−4) − 5(−1) =?
−19
−19 = −19 ✓

Section 5.2 Solving Systems of Linear Equations by Substitution
227
Solving a System of Linear Equations by Substitution
Solve the system of linear equations by substitution.
−x + y = 3 Equation 1
3x + y = −1 Equation 2
SOLUTION
Step 1 Solve for y in Equation 1.
y = x + 3 Revised Equation 1
Step 2 Substitute x + 3 for y in Equation 2 and solve for x.
3x + y = −1 Equation 2
3x + (x + 3) = −1 Substitute x + 3 for y.
4x + 3 = −1 Combine like terms.
4x = −4 Subtract 3 from each side.
x = −1 Divide each side by 4.
Step 3 Substitute −1 for x in Equation 1 and solve for y.
−x + y = 3 Equation 1
−(−1) + y = 3 Substitute −1 for x.
y = 2 Subtract 1 from each side.
The solution is (−1, 2).
Algebraic Check
Equation 1
−x + y = 3
−(−1) + 2 =?
3
3 = 3 ✓
Equation 2
3x + y = −1
3(−1) + 2 =?
−1
−1 = −1 ✓
Graphical Check
−5
−2
4
4
y = x + 3
4
y = −3x − 1
IntersectionX=1 Y=2
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.comSolve the system of linear equations by
substitution. Check your solution.
4. x + y = −2 5. −x + y = −4
−3x + y = 6 4x − y = 10
6. 2x − y = −5 7. x − 2y = 7
3x − y = 1 3x − 2y = 3
ANOTHER WAYYou could also begin by solving for x in Equation 1,
solving for y in Equation 2, or solving for x in Equation 2.

228 Chapter 5 Solving Systems of Linear Equations
Modeling with Mathematics
A drama club earns $1040 from a production. An adult ticket
costs twice as much as a student ticket. Write a system of linear
equations that represents this situation. What is the price of each
type of ticket?
SOLUTION
1. Understand the Problem You know the amount earned, the total
numbers of adult and student tickets sold, and the relationship
between the price of an adult ticket and the price of a student
ticket. You are asked to write a system of linear equations that
represents the situation and fi nd the price of each type of
ticket.
2. Make a Plan Use a verbal model to write a system of linear
equations that represents the problem. Then solve the system of
linear equations.
3. Solve the Problem
Words 64 ⋅ Adult ticket price + 132 ⋅ Student ticket price
= 1040
Adult ticket price
= 2 ⋅ Student ticket priceVariables Let x be the price (in
dollars) of an adult ticket and let y be the
price (in dollars) of a student ticket.
System 64x + 132y = 1040 Equation 1
x = 2y Equation 2
Step 1 Equation 2 is already solved for x.
Step 2 Substitute 2y for x in Equation 1 and solve for y.
64x + 132y = 1040 Equation 1
64(2y) + 132y = 1040 Substitute 2y for x.
260y = 1040 Simplify.
y = 4 Simplify.
Step 3 Substitute 4 for y in Equation 2 and solve for x.
x = 2y Equation 2
x = 2(4) Substitute 4 for y.
x = 8 Simplify.
The solution is (8, 4). So, an adult ticket costs $8 and a
student ticket costs $4.
4. Look Back To check that your solution is correct, substitute
the values of x and y into both of the original equations and
simplify.
64(8) + 132(4) = 1040 8 = 2(4)
1040 = 1040 ✓ 8 = 8 ✓
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.com 8. There are a total of 64 students in
a drama club and a yearbook club. The drama
club has 10 more students than the yearbook club. Write a system
of linear equations that represents this situation. How many
students are in each club?
Solving RealLife Problems
STUDY TIPYou can use either of the original equations to solve
for x. However, using Equation 2 requires fewer calculations.
AtsW
S
1
2
3
Tickets sold
Type Number
adult 64
student 132

Section 5.2 Solving Systems of Linear Equations by Substitution
229
Tutorial Help in English and Spanish at
BigIdeasMath.comExercises5.2
Monitoring Progress and Modeling with MathematicsMonitoring
Progress and Modeling with MathematicsIn Exercises 3−8, tell which
equation you would choose to solve for one of the variables.
Explain.
3. x + 4y = 30 4. 3x − y = 0 x − 2y = 0 2x + y = −10
5. 5x + 3y = 11 6. 3x − 2y = 19 5x − y = 5 x + y = 8
7. x − y = −3 8. 3x + 5y = 25 4x + 3y = −5 x − 2y = −6
In Exercises 9–16, solve the sytem of linear equations by
substitution. Check your solution. (See Examples 1 and 2.)
9. x = 17 − 4y 10. 6x − 9 = y y = x − 2 y = −3x
11. x = 16 − 4y 12. −5x + 3y = 51 3x + 4y = 8 y = 10x − 8
13. 2x = 12 14. 2x − y = 23 x − 5y = −29 x − 9 = −1
15. 5x + 2y = 9 16. 11x − 7y = −14 x + y = −3 x − 2y = −4
17. ERROR ANALYSIS Describe and correct the error in solving for
one of the variables in the linear system 8x + 2y = −12 and 5x − y
= 4.
Step 1 5x − y = 4 −y = −5x + 4 y = 5x − 4
Step 2 5x − (5x − 4) = 4 5x − 5x + 4 = 4 4 = 4
✗
18. ERROR ANALYSIS Describe and correct the error in solving for
one of the variables in the linear system 4x + 2y = 6 and 3x + y =
9.
Step 1 3x + y = 9 y = 9 − 3x
Step 2 4x + 2(9 − 3x) = 6 4x + 18 − 6x = 6 −2x = −12 x = 6
Step 3 3x + y = 9 3x + 6 = 9 3x = 3 x = 1
✗
19. MODELING WITH MATHEMATICS A farmer plants corn and wheat on
a 180acre farm. The farmer wants to plant three times as many
acres of corn as wheat. Write a system of linear equations that
represents this situation. How many acres of each crop should the
farmer plant? (See Example 3.)
20. MODELING WITH MATHEMATICS A company that offers tubing trips
down a river rents tubes for a person to use and “cooler” tubes to
carry food and water. A group spends $270 to rent a total of 15
tubes. Write a system of linear equations that represents this
situation. How many of each type of tube does the group rent?
1. WRITING Describe how to solve a system of linear equations by
substitution.
2. NUMBER SENSE When solving a system of linear equations by
substitution, how do you decide which variable to solve for in Step
1?
Vocabulary and Core Concept CheckVocabulary and Core Concept
Check

230 Chapter 5 Solving Systems of Linear Equations
Maintaining Mathematical ProficiencyMaintaining Mathematical
ProficiencyFind the sum or difference. (Skills Review Handbook)
36. (x − 4) + (2x − 7) 37. (5y − 12) + (−5y − 1)
38. (t − 8) − (t + 15) 39. (6d + 2) − (3d − 3)
40. 4(m + 2) + 3(6m − 4) 41. 2(5v + 6) − 6(−9v + 2)
Reviewing what you learned in previous grades and lessons
In Exercises 21–24, write a system of linear equations that has
the ordered pair as its solution.
21. (3, 5) 22. (−2, 8)
23. (−4, −12) 24. (15, −25)
25. PROBLEM SOLVING A math test is worth 100 points and has 38
problems. Each problem is worth either 5 points or 2 points. How
many problems of each point value are on the test?
26. PROBLEM SOLVING An investor owns shares of Stock A and Stock
B. The investor owns a total of 200 shares with a total value of
$4000. How many shares of each stock does the investor own?
Stock Price
A $9.50
B $27.00
MATHEMATICAL CONNECTIONS In Exercises 27 and 28, (a) write an
equation that represents the sum of the angle measures of the
triangle and (b) use your equation and the equation shown to fi nd
the values of x and y.
27.
x + 2 = 3y
x°
y°
28.
x°
y °(y − 18)°
3x − 5y = −22
29. REASONING Find the values of a and b so that the solution of
the linear system is (−9, 1).
ax + by = −31 Equation 1 ax − by = −41 Equation 2
30. MAKING AN ARGUMENT Your friend says that given a linear
system with an equation of a horizontal line and an equation of a
vertical line, you cannot solve the system by substitution. Is your
friend correct? Explain.
31. OPENENDED Write a system of linear equations in which (3,
−5) is a solution of Equation 1 but not a solution of Equation 2,
and (−1, 7) is a solution of the system.
32. HOW DO YOU SEE IT? The graphs of two linear equations are
shown.
2 4 6 x
2
4
6
y y = x + 1
y = 6 − x14
a. At what point do the lines appear to intersect?
b. Could you solve a system of linear equations by substitution
to check your answer in part (a)? Explain.
33. REPEATED REASONING A radio station plays a total of 272 pop,
rock, and hiphop songs during a day. The number of pop songs is 3
times the number of rock songs. The number of hiphop songs is 32
more than the number of rock songs. How many of each type of song
does the radio station play?
34. THOUGHT PROVOKING You have $2.65 in coins. Write a system of
equations that represents this situation. Use variables to
represent the number of each type of coin.
35. NUMBER SENSE The sum of the digits of a twodigit number is
11. When the digits are reversed, the number increases by 27. Find
the original number.

Section 5.3 Solving Systems of Linear Equations by Elimination
231
5.3 Solving Systems of Linear Equations by Elimination
Writing and Solving a System of Equations
Work with a partner. You purchase a drink and a sandwich for
$4.50. Your friend purchases a drink and fi ve sandwiches for
$16.50. You want to determine the price of a drink and the price of
a sandwich.
a. Let x represent the price (in dollars) of one drink. Let y
represent the price (in dollars) of one sandwich. Write a system of
equations for the situation. Use the following verbal model.
Number of drinks ⋅ Price per drink +
Number of sandwiches ⋅ Price per sandwich = Total price
Label one of the equations Equation 1 and the other equation
Equation 2.
b. Subtract Equation 1 from Equation 2. Explain how you can use
the result to solve the system of equations. Then fi nd and
interpret the solution.
Essential QuestionEssential Question How can you use elimination
to solve a system of linear equations?
Using Elimination to Solve a System
Work with a partner.
2x + y = 7 Equation 1 x + 5y = 17 Equation 2
a. Can you eliminate a variable by adding or subtracting the
equations as they are? If not, what do you need to do to one or
both equations so that you can?
b. Solve the system individually. Then exchange solutions with
your partner and compare and check the solutions.
Communicate Your AnswerCommunicate Your Answer 4. How can you
use elimination to solve a system of linear equations?
5. When can you add or subtract the equations in a system to
solve the system? When do you have to multiply fi rst? Justify your
answers with examples.
6. In Exploration 3, why can you multiply an equation in the
system by a constant and not change the solution of the system?
Explain your reasoning.
Using Elimination to Solve Systems
Work with a partner. Solve each system of linear equations using
two methods.
Method 1 Subtract. Subtract Equation 2 from Equation 1. Then use
the result to solve the system.
Method 2 Add. Add the two equations. Then use the result to
solve the system.
Is the solution the same using both methods? Which method do you
prefer?
a. 3x − y = 6 b. 2x + y = 6 c. x − 2y = −7
3x + y = 0 2x − y = 2 x + 2y = 5
USING PROBLEMSOLVING STRATEGIES
To be profi cient in math, you need to monitor and evaluate your
progress and change course using a different solution method, if
necessary.
A.2.IA.5.C
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS

232 Chapter 5 Solving Systems of Linear Equations
5.3 Lesson What You Will LearnWhat You Will Learn Solve systems
of linear equations by elimination.
Use systems of linear equations to solve reallife problems.
Solving Linear Systems by EliminationPreviouscoeffi cient
Core VocabularyCore Vocabullarry
Core Core ConceptConceptSolving a System of Linear Equations by
EliminationStep 1 Multiply, if necessary, one or both equations by
a constant so at least one
pair of like terms has the same or opposite coeffi cients.
Step 2 Add or subtract the equations to eliminate one of the
variables.
Step 3 Solve the resulting equation.
Step 4 Substitute the value from Step 3 into one of the original
equations and solve for the other variable.
Solving a System of Linear Equations by Elimination
Solve the system of linear equations by elimination.
3x + 2y = 4 Equation 13x − 2y = −4 Equation 2
SOLUTION
Step 1 Because the coeffi cients of the yterms are opposites,
you do not need to multiply either equation by a constant.
Step 2 Add the equations.
3x + 2y = 4 Equation 13x − 2y = −4 Equation 2
6x = 0 Add the equations.
Step 3 Solve for x.
6x = 0 Resulting equation from Step 2 x = 0 Divide each side by
6.
Step 4 Substitute 0 for x in one of the original equations and
solve for y.
3x + 2y = 4 Equation 13(0) + 2y = 4 Substitute 0 for x.
y = 2 Solve for y. The solution is (0, 2).
Check
Equation 1
3x + 2y = 4
3(0) + 2(2) =?
4
4 = 4 ✓Equation 2
3x − 2y = −4
3(0) − 2(2) =?
−4
−4 = −4 ✓
You can use elimination to solve a system of equations because
replacing one equation in the system with the sum of that equation
and a multiple of the other produces a system that has the same
solution. Here is why.
Consider System 1. In this system, a and c are algebraic
expressions, and b and d are constants. Begin by multiplying each
side of Equation 2 by a constant k. By the Multiplication Property
of Equality, kc = kd. You can rewrite Equation 1 as Equation 3 by
adding kc on the left and kd on the right. You can rewrite Equation
3 as Equation 1 by subtracting kc on the left and kd on the right.
Because you can rewrite either system as the other, System 1 and
System 2 have the same solution.
System 1
a = b Equation 1c = d Equation 2
System 2
a + kc = b + kd Equation 3c = d Equation 2

Section 5.3 Solving Systems of Linear Equations by Elimination
233
Solving a System of Linear Equations by Elimination
Solve the system of linear equations by elimination.
−10x + 3y = 1 Equation 1
−5x − 6y = 23 Equation 2
SOLUTION
Step 1 Multiply Equation 2 by −2 so that the coeffi cients of
the xterms are opposites.
−10x + 3y = 1 −10x + 3y = 1 Equation 1
−5x − 6y = 23 Multiply by −2. 10x + 12y = −46 Revised Equation
2
Step 2 Add the equations.
− 10x + 3y = 1 Equation 1 10x + 12y = −46 Revised Equation 2
15y = −45 Add the equations.
Step 3 Solve for y.
15y = −45 Resulting equation from Step 2y = −3 Divide each side
by 15.
Step 4 Substitute −3 for y in one of the original equations and
solve for x.
−5x − 6y = 23 Equation 2
−5x − 6(−3) = 23 Substitute −3 for y.
−5x + 18 = 23 Multiply.
−5x = 5 Subtract 18 from each side.
x = −1 Divide each side by −5.
The solution is (−1, −3).
Monitoring Progress Help in English and Spanish at
BigIdeasMath.comSolve the system of linear equations by
elimination. Check your solution.
1. 3x + 2y = 7 2. x − 3y = 24 3. x + 4y = 22
−3x + 4y = 5 3x + y = 12 4x + y = 13
ANOTHER WAYTo use subtraction toeliminate one of the variables,
multiply Equation 2 by 2 and then subtract the equations.
− 10x + 3y = 1−(−10x − 12y = 46) 15y = −45
Methods for Solving Systems of Linear Equations
Concept SummaryConcept Summary
Method When to Use
Graphing (Lesson 5.1) To estimate solutions
Substitution (Lesson 5.2)
When one of the variables in one of the equations has a coeffi
cient of 1 or −1
Elimination (Lesson 5.3)
When at least one pair of like terms has the same or opposite
coeffi cients
Elimination (Multiply First) (Lesson 5.3)
When one of the variables cannot be eliminated by adding or
subtracting the equations
Check
10
−10
−10
10
IntersectionX=1 Y=3
10
Equation 110
3
Equation 2

234 Chapter 5 Solving Systems of Linear Equations
Modeling with Mathematics
The graph represents the average salaries of classroom teachers
in two school districts. During what year were the average salaries
in the two districts equal? What was the average salary in both
districts in that year?
SOLUTION
1. Understand the Problem You know two points on each line in
the graph. You are asked to determine the year in which the average
salaries were equal and then determine the average salary in that
year.
2. Make a Plan Use the points in the graph to write a system of
linear equations. Then solve the system of linear equations.
3. Solve the Problem Find the slope of each line.
District A: m = 60 − 25 — 25 − 5
= 35 — 20
= 7 — 4 District B: m = 55 − 25 —
25 − 0 = 30 —
25 = 6 —
5
Use each slope and a point on each line to write equations of
the lines.
District A District B
y − y1 = m(x − x1) Write the pointslope form. y − y1 = m(x −
x1)
y − 25 = 7 — 4 (x − 5) Substitute for m, x1, and y1. y − 25
=
6 — 5 (x − 0)
−7x + 4y = 65 Write in standard form. −6x + 5y = 125
System −7x + 4y = 65 Equation 1
−6x + 5y = 125 Equation 2
Step 1 Multiply Equation 1 by −5. Multiply Equation 2 by 4.
−7x + 4y = 65 Multiply by −5. 35x − 20y = −325 Revised Equation
1
−6x + 5y = 125 Multiply by 4. −24x + 20y = 500 Revised Equation
2
Step 2 Add the equations.
35x − 20y = −325 Revised Equation 1
−24x + 20y = 500 Revised Equation 2
11x = 175 Add the equations.
Step 3 Solving the equation 11x = 175 gives x = 175 — 11 ≈
15.9.
Step 4 Substitute 175 — 11 for x in one of the original
equations and solve for y.
−7x + 4y = 65 Equation 1
−7 ( 175 — 11 ) + 4y = 65 Substitute 175 — 11 for x.y ≈ 44 Solve
for y.
The solution is about (15.9, 44). Because x ≈ 15.9 corresponds
to the year 2000, the average salary in both districts was about
$44,000 in 2000.
4. Look Back Using the graph, the point of intersection appears
to be about (15, 45). So, the solution of (15.9, 44) is
reasonable.
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.com 4. Write and solve a system of linear
equations represented by the graph at the left.
Solving RealLife Problems
STUDY TIPIn Example 3, both equations are multiplied by a
constant so that the coeffi cients of the yterms are
opposites.
00
20
40
10 20 30 x
y
Years since 1985
60
80
Classroom Teacher
Ave
rag
e sa
lary
(th
ou
san
ds
of
do
llars
) District ADistrict B
(25, 55)
(25, 60)
(5, 25)
(0, 25)
00
4
8
4 8 12 16 x
y
12
16
(2, 9)
(14, 15)
(4, 4)
(8, 16)Line B Line A

Section 5.3 Solving Systems of Linear Equations by Elimination
235
Tutorial Help in English and Spanish at
BigIdeasMath.comExercises5.3
Monitoring Progress and Modeling with MathematicsMonitoring
Progress and Modeling with MathematicsIn Exercises 3−10, solve the
system of linear equations by elimination. Check your solution.
(See Example 1.)
3. x + 2y = 13 4. 9x + y = 2 −x + y = 5 −4x − y = −17
5. 5x + 6y = 50 6. −x + y = 4 x − 6y = −26 x + 3y = 4
7. −3x − 5y = −7 8. 4x − 9y = −21 −4x + 5y = 14 −4x − 3y = 9
9. −y − 10 = 6x 10. 3x − 30 = y 5x + y = −10 7y − 6 = 3x
In Exercises 11–18, solve the system of linear equations by
elimination. Check your solution. (See Examples 2 and 3.)
11. x + y = 2 12. 8x − 5y = 11 2x + 7y = 9 4x − 3y = 5
13. 11x − 20y = 28 14. 10x − 9y = 46 3x + 4y = 36 −2x + 3y =
10
15. 4x − 3y = 8 16. −2x − 5y = 9 5x − 2y = −11 3x + 11y = 4
17. 9x + 2y = 39 18. 12x − 7y = −2 6x + 13y = −9 8x + 11y =
30
19. ERROR ANALYSIS Describe and correct the error in solving for
one of the variables in the linear system 5x − 7y = 16 and x + 7y =
8.
5x − 7y = 16 x + 7y = 8 4x = 24 x = 6
✗
20. ERROR ANALYSIS Describe and correct the error in solving for
one of the variables in the linear system 4x + 3y = 8 and x − 2y =
−13.
21. MODELING WITH MATHEMATICS A service center charges a fee of
x dollars for an oil change plus y dollars per quart of oil used. A
sample of its sales record is shown. Write a system of linear
equations that represents this situation. Find the fee and cost per
quart of oil.
A B
2
1
34
Customer Oil Tank Size(quarts)TotalCost
A 5B 7
C
$22.45$25.45
22. MODELING WITH MATHEMATICS The graph represents the average
salaries of high school principals in two states. During what year
were the average salaries in the two states equal? What was the
average salary in both states in that year?
1. OPENENDED Give an example of a system of linear equations
that can be solved by fi rst adding the equations to eliminate one
variable.
2. WRITING Explain how to solve the system of linear equations
2x − 3y = −4 Equation 1−5x + 9y = 7 Equation 2by elimination.
Vocabulary and Core Concept CheckVocabulary and Core Concept
Check
4x + 3y = 8 4x + 3y = 8
x − 2y = −13 Multiply by −4. −4x + 8y = −13
11y = −5
y = −5 — 11
✗
00
40
80
10 20 30 x
y
Years since 1985
120
160
High School Principal
Ave
rag
e sa
lary
(th
ou
san
ds
of
do
llars
)
State AState B
(22, 100)
(25, 98)(0, 43)
(6, 48)

236 Chapter 5 Solving Systems of Linear Equations
Maintaining Mathematical ProficiencyMaintaining Mathematical
ProficiencySolve the equation. Determine whether the equation has
one solution, no solution, or infi nitely many solutions. (Section
1.3)
36. 5d − 8 = 1 + 5d 37. 9 + 4t = 12 − 4t
38. 3n + 2 = 2(n − 3) 39. −3(4 − 2v) = 6v − 12
Write an equation of the line that passes through the given
point and is parallel to the given line. (Section 4.4)
40. (4, −1); y = −2x + 7 41. (0, 6); y = 5x − 3 42. (−5, −2); y
= 2 — 3 x + 1
Reviewing what you learned in previous grades and lessons
In Exercises 23–26, solve the system of linear equations using
any method. Explain why you chose the method.
23. 3x + 2y = 4 24. −6y + 2 = −4x 2y = 8 − 5x y − 2 = x
25. y − x = 2 26. 3x + y = 1 — 3
y = − 1 — 4 x + 7 2x − 3y = 8 — 3
27. WRITING For what values of a can you solve the linear system
ax + 3y = 2 and 4x + 5y = 6 by elimination without multiplying fi
rst? Explain.
28. HOW DO YOU SEE IT? The circle graph shows the results of a
survey in which 50 students were asked about their favorite
meal.
Favorite Meal
Dinner25
Breakfast
Lunch
a. Estimate the numbers of students who chose breakfast and
lunch.
b. The number of students who chose lunch was 5 more than the
number of students who chose breakfast. Write a system of linear
equations that represents the numbers of students who chose
breakfast and lunch.
c. Explain how you can solve the linear system in part (b) to
check your answers in part (a).
29. MAKING AN ARGUMENT Your friend says that any system of
equations that can be solved by elimination can be solved by
substitution in an equal or fewer number of steps. Is your friend
correct? Explain.
30. THOUGHT PROVOKING Write a system of linear equations that
can be added to eliminate a variable or subtracted to eliminate a
variable.
31. MATHEMATICAL CONNECTIONS A rectangle has a perimeter of 18
inches. A new rectangle is formed by doubling the width w and
tripling the lengthℓ, as shown. The new rectangle has a perimeter P
of 46 inches.
P = 46 in. 2w
3
a. Write and solve a system of linear equations to fi nd the
length and width of the original rectangle.
b. Find the length and width of the new rectangle.
32. CRITICAL THINKING Refer to the discussion of System 1 and
System 2 on page 232. Without solving, explain why the two systems
shown have the same solution.
System 1 System 2
3x − 2y = 8 Equation 1 5x = 20 Equation 3x + y = 6 Equation 2 x
+ y = 6 Equation 2
33. PROBLEM SOLVING You are making 6 quarts of fruit punch for a
party. You have bottles of 100% fruit juice and 20% fruit juice.
How many quarts of each type of juice should you mix to make 6
quarts of 80% fruit juice?
34. PROBLEM SOLVING A motorboat takes 40 minutes to travel 20
miles downstream. The return trip takes 60 minutes. What is the
speed of the current?
35. CRITICAL THINKING Solve for x, y, and z in the system of
equations. Explain your steps.
x + 7y + 3z = 29 Equation 1 3z + x − 2y = −7 Equation 2 5y = 10
− 2x Equation 3

Section 5.4 Solving Special Systems of Linear Equations 237
5.4 Solving Special Systems of Linear Equations
Using a Table to Solve a System
Work with a partner. You invest $450 for equipment to make
skateboards. The materials for each skateboard cost $20. You sell
each skateboard for $20.
a. Write the cost and revenue equations. Then copy and complete
the table for your cost C and your revenue R.
b. When will your company break even? What is wrong?
Essential QuestionEssential Question Can a system of linear
equations have no solution or infi nitely many solutions?
Writing and Analyzing a System
Work with a partner. A necklace and matching bracelet have two
types of beads. The necklace has 40 small beads and 6 large beads
and weighs 10 grams. The bracelet has 20 small beads and 3 large
beads and weighs 5 grams. The threads holding the beads have no
signifi cant weight.
a. Write a system of linear equations that represents the
situation. Let x be the weight (in grams) of a small bead and let y
be the weight (in grams) of a large bead.
b. Graph the system in the coordinate plane shown. What do you
notice about the two lines?
c. Can you fi nd the weight of each type of bead? Explain your
reasoning.
Communicate Your AnswerCommunicate Your Answer 3. Can a system
of linear equations have no solution or infi nitely many
solutions?
Give examples to support your answers.
4. Does the system of linear equations represented by each graph
have no solution, one solution, or infi nitely many solutions?
Explain.
a.
x
y
4
1
42−1
y = x + 2
x + y = 2
b.
x
y
3
6
42
3
y = x + 2
−x + y = 1
c.
x
y
3
1
6
42
3
y = x + 2
−2x + 2y = 4
x (skateboards) 0 1 2 3 4 5 6 7 8 9 10
C (dollars)
R (dollars)
APPLYING MATHEMATICSTo be profi cient in math, you need to
interpret mathematical results in reallife contexts.
x
y
1
1.5
2
0.5
00.2 0.3 0.40.10
A.2.IA.3.FA.5.C
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS

238 Chapter 5 Solving Systems of Linear Equations
5.4 Lesson What You Will LearnWhat You Will Learn Determine the
numbers of solutions of linear systems.
Use linear systems to solve reallife problems.
The Numbers of Solutions of Linear SystemsPreviousparallel
Core VocabularyCore Vocabullarry
Core Core ConceptConceptSolutions of Systems of Linear
EquationsA system of linear equations can have one solution, no
solution, or infi nitely many solutions.
One solution No solution Infi nitely many solutions
x
y
x
y
x
y
The lines intersect. The lines are parallel. The lines are the
same.
Solving a System: No Solution
Solve the system of linear equations.
y = 2x + 1 Equation 1y = 2x − 5 Equation 2
SOLUTION
Method 1 Solve by graphing.
Graph each equation.
The lines have the same slope and different yintercepts. So,
the lines are parallel.
Because parallel lines do not intersect, there is no point that
is a solution of both equations.
So, the system of linear equationshas no solution.
Method 2 Solve by substitution.
Substitute 2x − 5 for y in Equation 1.
y = 2x + 1 Equation 1
2x − 5 = 2x + 1 Substitute 2x − 5 for y.
−5 = 1 ✗ Subtract 2x from each side. The equation −5 = 1 is
never true. So, the system of linear equations
has no solution.
ANOTHER WAYYou can solve some linear systems by inspection. In
Example 1, notice you can rewrite the system as
–2x + y = 1–2x + y = –5.
This system has no solution because –2x + y cannot be equal to
both 1 and –5.
STUDY TIPA linear system with no solution is called an
inconsistent system.
x
y
2
−2
−4
41−21
2
1
2
2
−22
y = 2x + 1
x44
y = 2x − 5

Section 5.4 Solving Special Systems of Linear Equations 239
Solving a System: Infi nitely Many Solutions
Solve the system of linear equations.
−2x + y = 3 Equation 1
−4x + 2y = 6 Equation 2
SOLUTION
Solve by elimination.
Step 1 Multiply Equation 1 by −2.
−2x + y = 3 Multiply by −2. 4x − 2y = −6 Revised Equation 1
−4x + 2y = 6 −4x + 2y = 6 Equation 2
Step 2 Add the equations.
4x − 2y = −6 Revised Equation 1−4x + 2y = 6 Equation 2
0 = 0 Add the equations.
The equation 0 = 0 is always true. So, the solutions are all the
points on the line −2x + y = 3. The system of linear equations has
infi nitely many solutions.
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.comSolve the system of linear
equations.
1. x + y = 3 2. y = −x + 3
2x + 2y = 6 2x + 2y = 4
ANOTHER WAYYou can also solve the linear system by graphing. The
lines have the same slope and the same yintercept. So, the lines
are the same, which means all points on the line are solutions of
both equations.
STUDY TIPA linear system with infi nitely many solutions is
called a consistent dependent system.
Solving RealLife Problems
Modeling with Mathematics
An athletic director is comparing the costs of renting two
banquet halls for an awards banquet. Write a system of linear
equations that represents this situation. If the cost patterns
continue, will the cost of Hall A ever equal the cost of Hall
B?
SOLUTION
Words Total cost = Cost per hour ⋅ Number of hours + Initial
costVariables Let y be the cost (in dollars) and let x be the
number of hours.
System y = 100x + 75 Equation 1  Cost of Hall A
y = 100x + 100 Equation 2  Cost of Hall B
The equations are in slopeintercept form. The graphs of the
equations have the same slope but different yintercepts. There is
no solution because the lines are parallel.
So, the cost of Hall A will never equal the cost of Hall B.
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.com 3. WHAT IF? What happens to the
solution in Example 3 when the cost per hour for
Hall A is $125?
Cost (dollars)
Hours Hall A Hall B
0 75 100
1 175 200
2 275 300
3 375 400

240 Chapter 5 Solving Systems of Linear Equations
Modeling with Mathematics
The perimeter of the trapezoidal piece of land is 48 kilometers.
The perimeter of the rectangular piece of land is 144 kilometers.
Write and solve a system of linear equations to fi nd the values of
x and y.
SOLUTION
1. Understand the Problem You know the perimeter of each piece
of land and the side lengths in terms of x or y. You are asked to
write and solve a system of linear equations to fi nd the values of
x and y.
2. Make a Plan Use the fi gures and the defi nition of perimeter
to write a system of linear equations that represents the problem.
Then solve the system of linear equations.
3. Solve the Problem
Perimeter of trapezoid Perimeter of rectangle
2x + 4x + 6y + 6y = 48 9x + 9x + 18y + 18y = 144
6x + 12y = 48 Equation 1 18x + 36y = 144 Equation 2
System 6x + 12y = 48 Equation 1
18x + 36y = 144 Equation 2
Method 1 Solve by graphing.
Graph each equation.
The lines have the same slope and the same yintercept. So, the
lines are the same.
In this context, x and y must be positive. Because the lines are
the same, all the points on the line in Quadrant I are solutions of
both equations.
So, the system of linear equations has infi nitely many
solutions.
Method 2 Solve by elimination.
Multiply Equation 1 by −3 and add the equations.
6x + 12y = 48 Multiply by −3. −18x − 36y = −144 Revised Equation
1
18x + 36y = 144 18x + 36y = 144 Equation 2
0 = 0 Add the equations.
The equation 0 = 0 is always true. In this context, x and y must
be positive. So, the solutions are all the points on the line 6x +
12y = 48 in Quadrant I. The system of linear equations has infi
nitely many solutions.
4. Look Back Choose a few of the ordered pairs (x, y) that are
solutions of Equation 1. You should fi nd that no matter which
ordered pairs you choose, they will also be solutions of Equation
2. So, infi nitely many solutions seems reasonable.
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.com 4. WHAT IF? What happens to the
solution in Example 4 when the perimeter of the
trapezoidal piece of land is 96 kilometers? Explain.
4x
2x
6y 6y
18y
18y
9x9x
2 4 60 x
2
4
6
0
y
6x + 12y = 486x + 12y =
18x + 36y = 144

Section 5.4 Solving Special Systems of Linear Equations 241
Tutorial Help in English and Spanish at
BigIdeasMath.comExercises5.4
Monitoring Progress and Modeling with MathematicsMonitoring
Progress and Modeling with MathematicsIn Exercises 3−8, match the
system of linear equations with its graph. Then determine whether
the system has one solution, no solution, or infi nitely many
solutions.
3. −x + y = 1 4. 2x − 2y = 4 x − y = 1 −x + y = −2
5. 2x + y = 4 6. x − y = 0 −4x − 2y = −8 5x − 2y = 6
7. −2x + 4y = 1 8. 5x + 3y = 17 3x − 6y = 9 x − 3y = −2
A.
x
y
2
4
2−1
B.
x
y
2
4
6
1 4−2
C.
x
y
2
−2
2 4−2
D.
x
y
2
−3
1 4
E.
x
y
2
−1 2 4
F.
x
y
2
−3
2−3
In Exercises 9–16, solve the system of linear equations. (See
Examples 1 and 2.)
9. y = −2x − 4 10. y = −6x − 8 y = 2x − 4 y = −6x + 8
11. 3x − y = 6 12. −x + 2y = 7 −3x + y = −6 x − 2y = 7
13. 4x + 4y = −8 14. 15x − 5y = −20 −2x − 2y = 4 −3x + y = 4
15. 9x − 15y = 24 16. 3x − 2y = −5 6x − 10y = −16 4x + 5y =
47
In Exercises 17–22, use only the slopes and yintercepts of the
graphs of the equations to determine whether the system of linear
equations has one solution, no solution, or infi nitely many
solutions. Explain.
17. y = 7x + 13 18. y = −6x − 2 −21x + 3y = 39 12x + 2y = −6
19. 4x + 3y = 27 20. −7x + 7y = 1 4x − 3y = −27 2x − 2y =
−18
21. −18x + 6y = 24 22. 2x − 2y = 16 3x − y = −2 3x − 6y = 30
ERROR ANALYSIS In Exercises 23 and 24, describe and correct the
error in solving the system of linear equations.
23. −4x + y = 4 4x + y = 12
The lines do not intersect. So, the system has no solution.
✗x
y
1
−3
2−2
24. y = 3x − 8 y = 3x − 12
The lines have the same slope. So, the system has infi nitely
many solutions.
✗
1. REASONING Is it possible for a system of linear equations to
have exactly two solutions? Explain.
2. WRITING Compare the graph of a system of linear equations
that has infi nitely many solutions and the graph of a system of
linear equations that has no solution.
Vocabulary and Core Concept CheckVocabulary and Core Concept
Check

242 Chapter 5 Solving Systems of Linear Equations
Maintaining Mathematical ProficiencyMaintaining Mathematical
ProficiencySolve the system of linear equations by graphing.
(Section 5.1)
33. y = x − 6 34. y = 2x + 3 35. 2x + y = 6y = −x + 10 y = −3x −
7 3x − 2y = 16
Reviewing what you learned in previous grades and lessons
25. MODELING WITH MATHEMATICS The table shows the distances two
groups have traveled at different times during a canoeing
excursion. The groups continue traveling at their current rates
until they reach the same destination. Let d be the distance
traveled and t be the time since 1 p.m. Write a system of linear
equations that represents this situation. Will Group B catch up to
Group A before reaching the destination? Explain. (See Example
3.)
Distance Traveled (miles)
1 P.M. 2 P.M. 3 P.M. 4 P.M.
Group A 3 9 15 21
Group B 1 7 13 19
26. MODELING WITH MATHEMATICS A $6bag of trail mix contains 3
cups of dried fruit and 4 cups of almonds. A $9bag contains 4 1 —
2 cups of dried fruit and 6 cups of almonds. Write and solve a
system of linear equations to fi nd the price of 1 cup of dried
fruit and 1 cup of almonds. (See Example 4.)
27. PROBLEM SOLVING A train travels from New York City to
Washington, D.C., and then back to New York City. The table shows
the number of tickets purchased for each leg of the trip. The cost
per ticket is the same for each leg of the trip. Is there enough
information to determine the cost of one coach ticket? Explain.
DestinationCoach tickets
Business class
tickets
Money collected (dollars)
Washington, D.C. 150 80 22,860
New York City 170 100 27,280
28. THOUGHT PROVOKING Write a system of three linear equations
in two variables so that any two of the equations have exactly one
solution, but the entire system of equations has no solution.
29. REASONING In a system of linear equations, one equation has
a slope of 2 and the other equation has a slope of − 1 — 3 . How
many solutions does the system have? Explain.
30. HOW DO YOU SEE IT? The graph shows information about the
last leg of a 4 × 200meter relay for three relay teams. Team A’s
runner ran about 7.8 meters per second, Team B’s runner ran about
7.8 meters per second, and Team C’s runner ran about 8.8 meters per
second.
400
50
100
150
8 12 16 20 24 28
Dis
tan
ce (
met
ers)
x
y
Time (seconds)
Last Leg of 4 × 200Meter Relay
Team CTeam B
Team A
a. Estimate the distance at which Team C’s runner passed Team
B’s runner.
b. If the race was longer, could Team C’s runner have passed
Team A’s runner? Explain.
c. If the race was longer, could Team B’s runner have passed
Team A’s runner? Explain.
31. ABSTRACT REASONING Consider the system of linear equations y
= ax + 4 and y = bx − 2, where a and b are real numbers. Determine
whether each statement is always, sometimes, or never true. Explain
your reasoning.
a. The system has infi nitely many solutions.
b. The system has no solution.
c. When a < b, the system has one solution.
32. MAKING AN ARGUMENT One admission to an ice skating rink
costs x dollars, and renting a pair of ice skates costs y dollars.
Your friend says she can determine the exact cost of one admission
and one skate rental. Is your friend correct? Explain.
Total
2Admissions3
38.00
Skate Rentals
$ Total
15 Admissions10
190.00
Skate Rentals
$

243243
5.1–5.4 What Did You Learn?
Study Errors
What Happens: You do not study the right material or you do not
learn it well enough to remember it on a test without resources
such as notes.
How to Avoid This Error: Take a practice test. Work with a study
group. Discuss the topics on the test with your teacher. Do not try
to learn a whole chapter’s worth of material in one night.
Core VocabularyCore Vocabularysystem of linear equations, p. 220
solution of a system of linear equations, p. 220
Core ConceptsCore ConceptsSection 5.1Solving a System of Linear
Equations by Graphing, p. 221
Section 5.2Solving a System of Linear Equations by Substitution,
p. 226
Section 5.3Solving a System of Linear Equations by Elimination,
p. 232
Section 5.4Solutions of Systems of Linear Equations, p. 238
Mathematical ThinkingMathematical Thinking1. Describe the given
information in Exercise 33 on page 230 and your plan for fi nding
the solution.
2. Describe another reallife situation similar to Exercise 21
on page 235 and the mathematics that you can apply to solve the
problem.
3. What question(s) can you ask your friend to help her
understand the error in the statement she made in Exercise 32 on
page 242?
Study Skills
Analyzing Your Errors

244 Chapter 5 Solving Systems of Linear Equations
5.1–5.4 Quiz
Use the graph to solve the system of linear equations. Check
your solution. (Section 5.1)
1. y = − 1 — 3 x + 2 2. y = 1 — 2 x − 1 3. y = 1
y = x − 2 y = 4x + 6 y = 2x + 1
x
y3
1
−142−2
x
y2
−2
−4
−2−4
x
y
2
−2
2−2
Solve the system of linear equations by substitution. Check your
solution. (Section 5.2)
4. y = x − 4 5. 2y + x = −4 6. 3x − 5y = 13−2x + y = 18 y − x =
−5 x + 4y = 10
Solve the system of linear equations by elimination. Check your
solution. (Section 5.3)
7. x + y = 4 8. x + 3y = 1 9. 2x − 3y = −5−3x − y = −8 5x + 6y =
14 5x + 2y = 16
Solve the system of linear equations. (Section 5.4)
10. x − y = 1 11. 6x + 2y = 16 12. 3x − 3y = −2x − y = 6 2x − y
= 2 −6x + 6y = 4
13. You plant a spruce tree that grows 4 inches per year and a
hemlock tree that grows 6 inches per year. The initial heights are
shown. (Section 5.1)
a. Write a system of linear equations that represents this
situation.
b. Solve the system by graphing. Interpret your solution.
14. It takes you 3 hours to drive to a concert 135 miles away.
You drive 55 miles per hour on highways and 40 miles per hour on
the rest of the roads. (Section 5.1, Section 5.2, and Section
5.3)
a. How much time do you spend driving at each speed?
b. How many miles do you drive on highways? the rest of the
roads?
15. In a football game, all of the home team’s points are from
7point touchdowns and 3point fi eld goals. The team scores six
times. Write and solve a system of linear equations to fi nd the
numbers of touchdowns and fi eld goals that the home team scores.
(Section 5.1, Section 5.2, and Section 5.3)
sprucetree
hemlocktree
14 in.
8 in.

Section 5.5 Solving Equations by Graphing 245
Solving Equations by Graphing5.5
Solving an Equation by Graphing
Work with a partner. Solve 2x − 1 = − 1 — 2 x + 4 by
graphing.
a. Use the left side to write a linear equation. Then use the
right side to write another linear equation.
b. Graph the two linear equations from part (a). Find the
xvalue of the point of intersection. Check that the xvalue is the
solution of
2x − 1 = − 1 — 2 x + 4.
c. Explain why this “graphical method” works.
Essential QuestionEssential Question How can you use a system of
linear equations to solve an equation with variables on both
sides?
Previously, you learned how to use algebra to solve equations
with variables on both sides. Another way is to use a system of
linear equations.
Solving Equations Algebraically and Graphically
Work with a partner. Solve each equation using two methods.
Method 1 Use an algebraic method.
Method 2 Use a graphical method.
Is the solution the same using both methods?
a. 1 — 2 x + 4 = − 1 — 4 x + 1 b.
2 — 3 x + 4 =
1 — 3 x + 3
c. − 2 — 3 x − 1 = 1 — 3 x − 4 d.
4 — 5 x +
7 — 5 = 3x − 3
e. −x + 2.5 = 2x − 0.5 f. − 3x + 1.5 = x + 1.5
Communicate Your AnswerCommunicate Your Answer 3. How can you
use a system of linear equations to solve an equation with
variables on both sides?
4. Compare the algebraic method and the graphical method for
solving a linear equation with variables on both sides. Describe
the advantages and disadvantages of each method.
SELECTING TOOLSTo be profi cient in math, you need to consider
the available tools, which may include pencil and paper or a
graphing calculator, when solving a mathematical problem.
x
y
2
4
6
2 4 6−2
−2
A.5.A
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS

246 Chapter 5 Solving Systems of Linear Equations
5.5 Lesson What You Will LearnWhat You Will Learn Solve linear
equations by graphing. Use linear equations to solve reallife
problems.
Solving Linear Equations by GraphingYou can use a system of
linear equations to solve an equation with variables on both
sides.
Previousabsolute value equation
Core VocabularyCore Vocabullarry
Core Core ConceptConceptSolving Linear Equations by GraphingStep
1 To solve the equation ax + b = cx + d, write two linear
equations.
ax + b = cx + d
and
Step 2 Graph the system of linear equations. The xvalue of the
solution of the system of linear equations is the solution of the
equation ax + b = cx + d.
y = cx + dy = ax + b
Solving an Equation by Graphing
Solve −x + 1 = 2x − 5 by graphing. Check your solution.
SOLUTION
Step 1 Write a system of linear equations using each side of the
original equation.
−x + 1 = 2x − 5
Step 2 Graph the system.
y = −x + 1 Equation 1 y = 2x − 5 Equation 2
The graphs intersect at (2, −1).
So, the solution of the equation is x = 2.
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.comSolve the equation by graphing. Check
your solution.
1. 1 — 2 x − 3 = 2x 2. −4 + 9x = −3x + 2
Check
−x + 1 = 2x − 5
−(2) + 1 =?
2(2) − 5
−1 = −1 ✓
y = 2x − 5y = −x + 1
x
y1
1
y = −x + 1
y = 2x − 5
−1
−2
−4
(2, −1)

Section 5.5 Solving Equations by Graphing 247
Solving RealLife Problems
Modeling with Mathematics
You are an ecologist studying the populations of two types of fi
sh in a lake. Use the information in the table to predict when the
populations of the two types of fi sh will be equal.
SOLUTION
1. Understand the Problem You know the current population of
each type of fi sh and the annual change in each population. You
are asked to determine when the populations of the two types of fi
sh will be equal.
2. Make a Plan Use a verbal model to write an equation that
represents the problem. Then solve the equation by graphing.
3. Solve the Problem
Words Type A Type B
Change per year ⋅ Years +
Currentpopulation
= Change per year
⋅ Years + Currentpopulation Variable Let x be the number of
years.
Equation −250x + 5750 = 400x + 2825
Solve the equation by graphing.
Step 1 Write a system of linear equations using each side of the
original equation.
−250x + 5750 = 400x + 2825
Step 2 Graph the system.
y = −250x + 5750 Equation 1
y = 400x + 2825 Equation 2
The graphs appear to intersect at (4.5, 4625). Check this
solution in each equation of the linear system, as shown.
So, the populations of the two types of fi sh will be equal in
4.5 years.
4. Look Back To check that your solution is correct, verify that
x = 4.5 is the solution of the original equation.
−250(4.5) + 5750 = 400(4.5) + 2825
4625 = 4625 ✓
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.com 3. WHAT IF? Type C has a current
population of 3500 and grows by 500 each year.
Predict when the population of Type C will be equal to the
population of Type A.
Check
y = −250x + 5750
4625 =?
−250(4.5) + 5750
4625 = 4625 ✓
y = 400x + 2825
4625 =?
400(4.5) + 2825
4625 = 4625 ✓Years
Fish
po
pu
lati
on
x
y
2500
3500
4500
5500
6500
02 3 4 510
(0, 5750)
(4.5, 4625)
(0, 2825)
750)
(4.5, 4625)
y = −250x + 5750
))y = 400x + 2825
Fish in a Lake
TypeCurrent
populationChange per year
A 5750 −250
B 2825 400
y = 400x + 2825y = −250x + 5750

248 Chapter 5 Solving Systems of Linear Equations
Modeling with Mathematics
Your family needs to rent a car for a week while on vacation.
Company A charges $3.25 per mile plus a fl at fee of $125 per week.
Company B charges $3 per mile plus a fl at fee of $150 per week.
After how many miles of travel are the total costs the same at both
companies?
SOLUTION
1. Understand the Problem You know the costs of renting a car
from two companies. You are asked to determine how many miles of
travel will result in the same total costs at both companies.
2. Make a Plan Use a verbal model to write an equation that
represents the problem. Then solve the equation by graphing.
3. Solve the Problem
Words Company A Company B
Cost per mile ⋅ Miles +
Flatfee
= Cost per mile
⋅ Miles + FlatfeeVariable Let x be the number of miles
traveled.
Equation 3.25x + 125 = 3x + 150
Solve the equation by graphing.
Step 1 Write a system of linear equations using each side of the
original equation.
3.25x + 125 = 3x + 150
Step 2 Use a graphing calculator to graph the system.
00
600
150IntersectionX=100 Y=450
ction
y = 3.25x + 125
y = 3x + 150
Because the graphs intersect at (100, 450), the solution of the
equation is x = 100.
So, the total costs are the same after 100 miles.
4. Look Back One way to check your solution is to solve the
equation algebraically, as shown.
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.com 4. WHAT IF? Company C charges $3.30 per
mile plus a fl at fee of $115 per
week. After how many miles are the total costs the same at
Company A and Company C?
y = 3x + 150y = 3.25x + 125
Check
3.25x + 125 = 3x + 150 0.25x + 125 = 150 0.25x = 25 x = 100

Section 5.5 Solving Equations by Graphing 249
Tutorial Help in English and Spanish at
BigIdeasMath.comExercises5.5
Monitoring Progress and Modeling with MathematicsMonitoring
Progress and Modeling with MathematicsIn Exercises 3–10, use the
graph to solve the equation. Check your solution.
3. x − 3 = 2 4. −3 = 4x + 1
x
y
−2
1
3
1 4
x
y
1
2−2
5. −2x + 3 = x 6. x − 4 = 3x
x
y
1
3
1 3−3
xy
−6
2−2
7. 2x + 1 = −x − 2 8. −5x − 1 = 3x − 1
x
y
−3
2
2−2
x
y
−1 2−2
9. −x − 1 = 1 — 3 x + 3 10. − 3 — 2 x − 2 = −4x + 3
x
y
2
4
1−2−4
x
y
2 4
−2
−4
−6
In Exercises 11−18, solve the equation by graphing. Check your
solution. (See Example 1.)
11. x + 4 = −x 12. 4x = x + 3
13. x + 5 = −2x − 4 14. −2x + 6 = 5x − 1
15. 1 — 2 x − 2 = 9 − 5x 16. −5 + 1 — 4 x = 3x + 6
17. 5x − 7 = 2(x + 1) 18. −6(x + 4) = −3x − 6
In Exercises 19−24, solve the equation by graphing. Determine
whether the equation has one solution, no solution, or infi nitely
many solutions.
19. 3x − 1 = −x + 7 20. 5x − 4 = 5x + 1
21. −4(2 − x) = 4x − 8
22. −2x − 3 = 2(x − 2)
23. −x − 5 = − 1 — 3 (3x + 5)
24. 1 — 2 (8x + 3) = 4x + 3 — 2
In Exercises 25–28, write an equation that has the same solution
as the linear system represented by the graph.
25.
x
y
−2
2
4
2 4
26.
x
y
2
4
2 4
27.
x
y
−4
2
2−4
28.
x
y
2
−2
4
1. REASONING The graphs of the equations y = 3x − 20 and y = −2x
+ 10 intersect at the point (6, −2). Without solving, fi nd the
solution of the equation 3x − 20 = −2x + 10.
2. WRITING Consider the equation ax + b = cx + d, where c = 0.
Can you solve the equation by graphing a system of linear
equations? Explain.
Vocabulary and Core Concept CheckVocabulary and Core Concept
Check

250 Chapter 5 Solving Systems of Linear Equations
Maintaining Mathematical ProficiencyMaintaining Mathematical
ProficiencyGraph the inequality. (Section 2.1)
40. y > 5 41. x ≤ −2 42. n ≥ 9 43. c < −6
Use the graphs of f and g to describe the transformation from
the graph of f to the graph of g. (Section 3.7)
44. f(x) = x − 5; g(x) = f(x + 2) 45. f(x) = 6x; g(x) =
−f(x)
46. f(x) = −2x + 1; g(x) = f(4x) 47. f(x) = 1 — 2 x − 2; g(x) =
f(x − 1)
Reviewing what you learned in previous grades and lessons
USING TOOLS In Exercises 29 and 30, use a graphing calculator to
solve the equation.
29. 0.7x + 0.5 = −0.2x − 1.3
30. 2.1x + 0.6 = −1.4x + 6.9
31. MODELING WITH MATHEMATICS You need to hire a catering
company to serve meals to guests at a wedding reception. Company A
charges $500 plus $20 per guest. Company B charges $800 plus $16
per guest. For how many guests are the total costs the same at both
companies? (See Examples 2 and 3.)
32. MODELING WITH MATHEMATICS Your dog is 16 years old in dog
years. Your cat is 28 years old in cat years. For every human year,
your dog ages by 7 dog years and your cat ages by 4 cat years. In
how many human years will both pets be the same age in their
respective types of years?
33. MODELING WITH MATHEMATICS You and a friend race across a fi
eld to a fence. Your friend has a 50meter head start. The
equations shown represent you and your friend’s distances d (in
meters) from the fence t seconds after the race begins. Find the
time at which you catch up to your friend.
You: d = −5t + 200
Your friend: d = −3 1 — 3 t + 150
34. MAKING AN ARGUMENT The graphs of y = −x + 4 and y = 2x − 8
intersect at the point (4, 0). So, your friend says the solution of
the equation −x + 4 = 2x − 8 is (4, 0). Is your friend correct?
Explain.
35. OPENENDED Find values for m and b so that the solution of
the equation mx + b = − 2x − 1 is x = −3.
36. HOW DO YOU SEE IT? The graph shows the total revenue and
expenses of a company x years after it opens for business.
200
2
4
6
4 6 8 10
Mill
ion
s o
f d
olla
rs
x
y
Year
Revenue and Expenses
revenue
expenses
a. Estimate the point of intersection of the graphs.
b. Interpret your answer in part (a).
37. MATHEMATICAL CONNECTIONS The value of the perimeter of the
triangle (in feet) is equal to the value of the area of the
triangle (in square feet). Use a graph to fi nd x.
38. THOUGHT PROVOKING A car has an initial value of $20,000 and
decreases in value at a rate of $1500 per year. Describe a
different car that will be worth the same amount as this car in
exactly 5 years. Specify the initial value and the rate at which
the value decreases.
39. ABSTRACT REASONING Use a graph to determine the sign of the
solution of the equation ax + b = cx + d in each situation.
a. 0 < b < d and a < c b. d < b < 0 and a <
c
6 ftx ft
(x − 2) ft

Section 5.6 Linear Inequalities in Two Variables 251
5.6 Linear Inequalities in Two Variables
Essential QuestionEssential Question How can you write and graph
a linear inequality in two variables?
A solution of a linear inequality in two variables is an ordered
pair (x, y) that makes the inequality true. The graph of a linear
inequality in two variables shows all the solutions of the
inequality in a coordinate plane.
Writing a Linear Inequality in Two Variables
Work with a partner.
a. Write an equation represented by the dashed line.
b. The solutions of an inequality are represented by the shaded
region. In words, describe the solutions of the inequality.
c. Write an inequality represented by the graph. Which
inequality symbol did you use? Explain your reasoning.
Graphing Linear Inequalities in Two Variables
Work with a partner. Graph each linear inequality in two
variables. Explain your steps. Use a graphing calculator to check
your graphs.
a. y > x + 5 b. y ≤ − 1 — 2 x + 1 c. y ≥ −x − 5
Communicate Your AnswerCommunicate Your Answer 4. How can you
write and graph a linear inequality in two variables?
5. Give an example of a reallife situation that can be modeled
using a linear inequality in two variables.
SELECTING TOOLSTo be profi cient in math, you need to use
technological tools to explore and deepen your understanding of
concepts.
x
y
2
4
2 4−2−4
−2
Using a Graphing Calculator
Work with a partner. Use a graphing calculator to graph y ≥ 1 —
4 x − 3.
a. Enter the equation y = 1 — 4 x − 3 into your calculator.
b. The inequality has the symbol ≥. So, the region to be shaded
is above the graph of y = 1 — 4 x − 3, as shown. Verify this by
testing a point in this region, such as (0, 0), to make sure it is
a solution of the inequality.
Because the inequality symbol is greater than or equal to, the
line is solid and not dashed. Some graphing calculators always use
a solid line when graphing inequalities. In this case, you have to
determine whether the line should be solid or dashed, based on the
inequality symbol used in the original inequality.
10
−10
−10
10
y ≥ x − 314
A.2.HA.3.D
TEXAS ESSENTIAL KNOWLEDGE AND SKILLS

252 Chapter 5 Solving Systems of Linear Equations
5.6 Lesson What You Will LearnWhat You Will Learn Check
solutions of linear inequalities. Graph linear inequalities in two
variables.
Write linear inequalities in two variables.
Use linear inequalities to solve reallife problems.
Linear InequalitiesA linear inequality in two variables, x and
y, can be written as
ax + by < c ax + by ≤ c ax + by > c ax + by ≥ c
where a, b, and c are real numbers. A solution of a linear
inequality in two variables is an ordered pair (x, y) that makes
the inequality true.
linear inequality in two variables, p. 252solution of a linear
inequality in two variables, p. 252graph of a linear inequality, p.
252halfplanes, p. 252
Previousordered pair
Core VocabularyCore Vocabullarry
Checking Solutions
Tell whether the ordered pair is a solution of the
inequality.
a. 2x + y < −3; (−1, 9) b. x − 3y ≥ 8; (2, −2)
SOLUTION
a. 2x + y < −3 Write the inequality.
2(−1) + 9 <?
−3 Substitute −1 for x and 9 for y.
7 < −3 ✗ Simplify. 7 is not less than −3. So, (−1, 9) is not
a solution of the inequality.
b. x − 3y ≥ 8 Write the inequality.
2 − 3(−2) ≥?
8 Substitute 2 for x and −2 for y.
8 ≥ 8 ✓ Simplify. 8 is equal to 8. So, (2, −2) is a solution of
the inequality.
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.comTell whether the ordered pair is a
solution of the inequality.
1. x + y > 0; (−2, 2) 2. 4x − y ≥ 5; (0, 0)
3. 5x − 2y ≤ −1; (−4, −1) 4. −2x − 3y < 15; (5, −7)
Graphing Linear Inequalities in Two VariablesThe graph of a
linear inequality in two variables shows all the solutions of the
inequality in a coordinate plane.
x
y4
2
2−2
All solutions of < 2lie on one side of the = 2 .
The boundary line dividesthe coordinate plane into
twohalfplanes. The shadedhalfplane is the graph of < 2 .
y
y
x
xy x
boundary line
READINGA dashed boundary line means that points on the line are
not solutions. A solid boundary line means that points on the line
are solutions.

Section 5.6 Linear Inequalities in Two Variables 253
Graphing a Linear Inequality in One Variable
Graph y ≤ 2 in a coordinate plane.
SOLUTION
Step 1 Graph y = 2. Use a solid line because the
x
y
1
3
2 4−1
(0, 0)
inequality symbol is ≤.
Step 2 Test (0, 0).
y ≤ 2 Write the inequality.
0 ≤ 2 ✓ Substitute.Step 3 Because (0, 0) is a solution, shade
the
halfplane that contains (0, 0).
Check
3
−1
−2
5
Core Core ConceptConceptGraphing a Linear Inequality in Two
VariablesStep 1 Graph the boundary line for the inequality. Use a
dashed line for < or >.
Use a solid line for ≤ or ≥.
Step 2 Test a point that is not on the boundary line to
determine whether it is a solution of the inequality.
Step 3 When the test point is a solution, shade the halfplane
that contains the point. When the test point is not a solution,
shade the halfplane that does not contain the point.
Graphing a Linear Inequality in Two Variables
Graph −x + 2y > 2 in a coordinate plane.
SOLUTION
Step 1 Graph −x + 2y = 2, or y = 1 — 2 x + 1. Use a
x
y
2
4
2−2
(0, 0)
dashed line because the inequality symbol is >.
Step 2 Test (0, 0).
−x + 2y > 2 Write the inequality.
−(0) + 2(0) >?
2 Substitute.
0 > 2 ✗ Simplify.Step 3 Because (0, 0) is not a solution,
shade the
halfplane that does not contain (0, 0).
Monitoring ProgressMonitoring Progress Help in English and
Spanish at BigIdeasMath.comGraph the inequality in a coordinate
plane.
5. y > −1 6. x ≤ −4
7. x + y ≤ −4 8. x − 2y < 0
STUDY TIPIt is often convenient to use the origin as a test
point. However, you must choose a different test point when the
origin is on the boundary line.

254 Chapter 5 Solving Systems of Linear Equations
Writing Linear Inequalities in Two Variables
Core Core ConceptConceptWriting a Linear Inequality in Two
Variables Using a GraphWrite an equation in slopeintercept form of
the boundary line.
If the shaded halfplane is above the boundary line, then
• replace = with > when the boundary line is dashed.
• replace = with ≥ when the boundary line is solid.
If the shaded halfplane is below the boundary line, then
• replace = with < when the boundary line is dashed.
• replace = with ≤ when the boundary line is solid.
Writing a Linear Inequality Using a Graph
Write an inequality that represents the graph.
SOLUTION
The boundar